Example – 9

Consider three points on the ellipse \(\begin{align}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\,,\,\,P({\theta _1}),\,\,Q({\theta _2})\;and\;R({\theta _3}).\end{align}\) What is the area of \(\Delta PQR\,\,?\) When is this area maximum?

Solution: The three points have the coordinates

\[P \equiv (a\cos {\theta _1},\,\,b\sin {\theta _1});\,\,\,Q \equiv (a\cos {\theta _2},b\sin {\theta _2});\,\,\,R \equiv (a\cos {\theta _3},\,b\sin {\theta _3})\]

The area of this triangle, by the determinant formula, is

\[\Delta = \frac{1}{2}\left| {\begin{align}&{a\cos {\theta _1}}&&{b\sin {\theta _1}}&&1\\&{a\cos {\theta _2}}&&{b\sin {\theta _2}}&&1\\&{a\cos {\theta _3}}&&{b\sin {\theta _3}}&&1\end{align}} \right|\quad\quad\quad...\left( 1 \right)\]

\[\begin{align}& = \frac{{ab}}{2}\left\{ {\cos {\theta _1}(\sin {\theta _2} - \sin {\theta _3}) + \sin {\theta _1}(\cos {\theta _3} - \cos {\theta _2}) + \left( {\cos {\theta _2}\sin {\theta _3} - \sin {\theta _2}\cos {\theta _3}} \right)} \right\}\\ &= \frac{{ab}}{2}\left\{ {\sin ({\theta _2} - {\theta _1}) + \sin ({\theta _1} - {\theta _3}) + \sin ({\theta _3} - {\theta _2})} \right\}\\ &= \frac{{ab}}{2}\left\{ {\sin ({\theta _2} - {\theta _1}) + 2\sin \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right)\cos \left( {\frac{{{\theta _1} + {\theta _2}}}{2} - {\theta _3}} \right)} \right\}\\ &= ab\sin \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right)\left\{ {\cos \left( {{\theta _3} - \frac{{{\theta _1} + {\theta _2}}}{2}} \right) - \cos \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right)} \right\}\\ &= 2ab\sin \left( {\frac{{{\theta _1} - {\theta _2}}}{2}} \right)\sin \left( {\frac{{{\theta _2} - {\theta _3}}}{2}} \right)\sin \left( {\frac{{{\theta _3} - {\theta _1}}}{2}} \right)\end{align}\]

This is the area of the triangle PQR.

To find its maximum value, we use a rather indirect route. Suppose we had to calculate the area \(\Delta '\) of a triangle inscribed in the circle \({x^2} + {y^2} = {a^2}\) with the same polar angles as P, Q, R. The only difference between \(\Delta '\;and\;\Delta \) will be that in the determinant expression for \(\Delta \) in (1), we will have all ‘a’ instead of ‘b’ in the terms of the second column.

This means that \(\Delta \;and\;\Delta '\) will always be in a constant ratio:

\[\frac{\Delta }{{\Delta '}} = \frac{b}{a}\]

Thus, the maximum for \(\Delta \) will be achieved in the same configuration as the one in which the maximum of \(\Delta '\) be achieved!

Since the area of a triangle inscribed in a circle has the maximum value when that triangle is equilateral (this should be intuitively obvious but can also be easily proved), \(\Delta '\) and hence \(\Delta \) will

be maximum when

\[\left| {{\theta _1} - {\theta _2}} \right| = \left| {{\theta _2} - {\theta _3}} \right| = \left| {{\theta _3} - {\theta _1}} \right| = \frac{{2\pi }}{3}\]

Thus, the three eccentric angels must be equally spaced apart at \(\frac{{2\pi }}{3}.\)

Example – 10

Prove that the circle on any focal distance as diameter touches the auxiliary circle of the ellipse.

Solution: Let \(P(\theta )\) be an arbitrary point on the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) and let F1 be one of its foci.

The radius of the auxiliary is a. The circle on PF1 as diameter will touch the auxiliary circle (internally) if:

\[OC + ({\rm{radius\; of\; this\; circle}}) = a\]

C, being the mid-point of PF1, has the coordinates

\[C \equiv \left( {\frac{{a\cos \theta - ae}}{2},\,\,\frac{{b\sin \theta }}{2}} \right)\]

Thus,

\[\begin{align}&OC = \sqrt {{{\left( {\frac{{a\cos \theta - ae}}{2}} \right)}^2} + {{\left( {\frac{{b\sin \theta }}{2}} \right)}^2}} \\&\quad~{\rm{ }} = \frac{1}{2}\sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta + {a^2}{e^2} - 2{a^2}e\cos \theta } \\&\quad~{\rm{ }} = \frac{a}{2}\sqrt {{{\cos }^2}\theta + (1 - {e^2}){{\sin }^2}\theta + {e^2} - 2e\cos \theta } \\&\quad{\rm{ }} = \frac{a}{2}\sqrt {1 + {e^2}{{\cos }^2}\theta - 2e\cos \theta } = \frac{a}{2}(1 - e\cos \theta )\end{align}\]

Also, the radius of the inner circle is

\[\begin{align}&C{F_1} \equiv \sqrt {{{\left( {\frac{{a\cos \theta - ae}}{2} + ae} \right)}^2} + {{\left( {\frac{{b\sin \theta }}{2}} \right)}^2}} \\&\quad~~~{\rm{ }} = \frac{a}{2}\sqrt {{{\cos }^2}\theta + (1 - {e^2}){{\sin }^2}\theta + {e^2} + 2\cos \theta } = \frac{a}{2}(1 + e\cos \theta )\end{align}\]

This gives

\(OC + C{F_1} = a\) which proves the stated assertion.

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